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91Ó°ÊÓ

A random sample of 49 measurements from one population had a sample mean of 10 , with sample standard deviation 3 . An independent random sample of 64 measurements from a second population had a sample mean of 12, with sample standard deviation \(4 .\) Test the claim that the population means are different. Use level of significance \(0.01\). (a) What distribution does the sample test statistic follow? Explain. (b) State the hypotheses. (c) Compute \(\bar{x}_{1}-\bar{x}_{2}\) and the corresponding sample test statistic. (d) Estimate the \(P\) -value of the sample test statistic. (e) Conclude the test. (f) The results.

Short Answer

Expert verified
The population means are significantly different, as the P-value is less than 0.01, leading to the rejection of the null hypothesis.

Step by step solution

01

Distribution of the Sample Test Statistic

Since both sample sizes are large (greater than 30) and the population standard deviations are unknown, the sample test statistic follows a normal distribution according to the central limit theorem. Specifically, the distribution used here is a standard normal (Z) distribution.
02

State the Hypotheses

We need to test whether the means of the two populations are different. The null hypothesis (H0) is that the means are equal: \( H_0: \mu_1 = \mu_2 \). The alternative hypothesis (H1) is that the means are not equal: \( H_1: \mu_1 eq \mu_2 \).
03

Calculate the Difference in Sample Means

The difference in sample means is given by \( \bar{x}_1 - \bar{x}_2 = 10 - 12 = -2 \).
04

Compute the Test Statistic

The test statistic is calculated using the formula for the standard normal Z-test for two means:\[Z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]Since \( \mu_1 - \mu_2 = 0 \) under the null hypothesis, we have:\[Z = \frac{-2 - 0}{\sqrt{\frac{3^2}{49} + \frac{4^2}{64}}} = \frac{-2}{\sqrt{\frac{9}{49} + \frac{16}{64}}}\]Calculating further:\[Z = \frac{-2}{\sqrt{0.1847}} = \frac{-2}{0.4297} \approx -4.66\]
05

Estimate the P-value

Using the standard normal distribution, find the P-value corresponding to the calculated Z-value of approximately -4.66. Since this is a two-tailed test, determine the probability in both tails of the distribution. Using a Z-table or calculator, the P-value is very small (much less than 0.01).
06

Conclude the Test

Since the P-value is less than the significance level of 0.01, we reject the null hypothesis. There is sufficient evidence to conclude that the population means are different.
07

Results Summary

The test showed a significant difference between the population means, as the P-value was less than the significance level of 0.01, suggesting that the means are not equal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Z-test
A Z-test is a statistical method used to determine if there is a significant difference between the means of two independent samples. It's particularly useful when the sample sizes are large, typically greater than 30. The Z-test calculates a test statistic based on the difference between the sample means, under the assumption that the difference follows a standard normal distribution.

For our specific example, the Z-test is applied to compare the means of two different populations. Given the sample sizes of 49 and 64, both larger than 30, the Z-test is suitable because it enables us to use the properties of the normal distribution to make inferences about the population means.

Key steps in performing a Z-test include:
  • Calculating the difference between the sample means
  • Determining the standard error of this difference
  • Computing the Z-test statistic using the formula: \[Z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]
The resulting Z-value helps us decide whether to reject the null hypothesis of equal population means.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental concept in statistics. It states that the sampling distribution of the sample mean will tend to be normal or nearly normal if the sample size is large enough, irrespective of the shape of the population distribution.

In our exercise, since both sample sizes are greater than 30, the CLT assures us that the distribution of the sample test statistic is approximately normal. This is critical because it validates our use of the Z-test. By applying the CLT, we're able to use a normal distribution to make inferences about our population means, even if we don't know the true population standard deviations.

Here’s why the CLT is important:
  • It allows for the approximation of the mean of any population as a normal distribution when the sample size is sufficiently large.
  • It lays the foundation for hypothesis testing and inferential statistics.
  • The theorem supports the assumption that averages of samples have a reliable pattern of variability.
Ultimately, the CLT is a powerful tool that underpins much of statistical theory and practice.
P-value
The P-value is a crucial concept in hypothesis testing, representing the probability of observing your results, or something more extreme, if the null hypothesis is true. Essentially, the P-value quantifies the evidence against the null hypothesis.

In the context of our exercise, we calculated a Z-test statistic of approximately -4.66. By referencing a standard normal distribution table, we estimate the P-value associated with this statistic. For a two-tailed test like ours, this involves finding the probability in both tails, relevant because we're testing for any difference in population means, not just an increase or decrease.

To interpret the P-value:
  • A small P-value (typically ≤ 0.05) indicates strong evidence against the null hypothesis, so we reject it.
  • A large P-value (> 0.05) suggests weak evidence against the null hypothesis, so we fail to reject it.
  • In this exercise, the P-value is much smaller than 0.01, meaning strong evidence exists to support the claim that the means are different.
Understanding P-values helps in making informed decisions about the validity of our hypotheses.
Significance Level
The significance level, often denoted as \( \alpha \), is a threshold set by the researcher that determines when the null hypothesis should be rejected. It is usually set at common levels such as 0.01, 0.05, or 0.10, depending on the desired balance between Type I and Type II errors.

In our test, the significance level is set at 0.01. This is a stringent criterion, meaning we require strong evidence (a very low P-value) to conclude that the population means are different. Using a 0.01 significance level reduces the likelihood of incorrectly rejecting the null hypothesis when it is true (Type I error).

Key aspects of setting a significance level:
  • It reflects tolerance for Type I error—rejecting a true null hypothesis.
  • A lower \( \alpha \) indicates a stricter criterion, demanding more evidence to reject the null.
  • This level should be set before conducting the test to avoid bias in decision-making.
In conclusion, the significance level is a critical part of hypothesis testing, as it guides the decision-making process.

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